Bài 2:
Theo đề ta có: \(a,b>0\left(1\right)\)
Giả sử: \(a^2+b^2\ge1\left(2\right)\)
Từ: \(\left(1\right)\left(2\right)\Rightarrow a^3+b^3\le\left(a-b\right)\left(a^2+b^2\right)\)
\(\Leftrightarrow a^3+b^3\le a^3+ab^2-a^2b-b^3\)
\(\Leftrightarrow ab^2-a^2b-2b^3\ge0\)
\(\Leftrightarrow b\left(ab-a^2-2b^2\right)\ge0\)
Vì: \(a>b>0\Rightarrow a\left(b-a\right)< 0\)
\(\Rightarrow a\left(b-a\right)-2b^3< 0\)
Từ trên suy ra BĐT trên không thể xảy ra.
\(\RightarrowĐpcm\)
Bài 1:
\(x^2+2x+3-\left(2x+1\right)\sqrt{x^2+2x+3}+4x-2=0\)
Đặt \(\sqrt{x^2+2x+3}=a>0\)
\(\Leftrightarrow a^2-\left(2x+1\right)a+4x-2=0\)
\(\Delta=\left(2x+1\right)^2-4\left(4x-2\right)=4x^2-12x+9=\left(2x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{2x+1-\left(2x-3\right)}{2}=2\\a=\frac{2x+1+2x-3}{2}=2x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x+3}=2\\\sqrt{x^2+2x+3}=2x-1\left(x\ge\frac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+3=4\\x^2+2x+3=4x^2-4x+1\end{matrix}\right.\)
\(\Leftrightarrow...\)
